Question

PLASE HELP WILL MEDAL AND FAN!!
Solve for n: 21k – 3n + 9p > 3p + 12.

A. n > 2p + 7k – 4
B. n < 2p + 7k – 4
C. n > –4p – 7k + 4
D. n < –4p – 7k + 4

Answer 1

Start by adding 3p to both sides
21k−3n+9p+3p>3p+12+3p
What is the expression we are left with?

Answer 2

idk i dont get the problem at all @pooja195

Answer 3

Alright
Let's go through this step by step :)
We want p on one side corrcect?
Good so to start off we need to start by moving all the p's to one side to do this we should start by canceling what we can
21k – 3n + 9p > 3p + 12.
Start by subtract p fro both sides
3-3=0
9-3=?

Answer 4

6

Answer 5

good now we have
21k−3n+6p>12
Any ideas on what the next step might be/

Answer 6

*?

Answer 7

no

Answer 8

Ok we are isolating p so we need to subtract 21k to both sides

Answer 9

21k-21k−3n+6p>12-21k
What are we left with?

Answer 10

wait we are solving for n ....

Answer 11

im so sorry wait im gonna rewrite the steps
Solve for n: 21k – 3n + 9p > 3p + 12.
Subtract 9p
21k - 3n + 9p > 3p + 12
subtract 21k
-3n > -6p - 21k + 12

Answer 12

Any ideas on what to do from here??

Answer 13

subtract 6p?

Answer 14

hmm nope we need to isolate the n so just divide everything by -3
-3/-3=n
-6/-3=?
-21/-3=?
12/-3=?

Answer 15

2
7
-4

Answer 16

Well done ^_^
now we have
n>7k+2p−4
HOWEVER when we divide by a negative we FLIP the sign so it would be
n<7k+2p−4

Answer 17

Do you understand the concept? :)

Answer 18

I hope i didnt confuse you too much

Answer 19

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Answer 20

yes thanks

Answer 21

=)